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Math Help - Unity in Subrings, Zero Divisors

  1. #1
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    Unity in Subrings, Zero Divisors

    Hello!

    I have two questions!

    (1)

    if M is a subring of N, where N has unity (let's call it 1_N), and say 1_M is the unity of M. Now, I have shown that it is NOT necessarily true that 1_M = 1_N

    Then if M and N are both fields (every element is a unit) then I'm trying to show that it is necessarily true that 1_M = 1_N.

    I am thinking you just take any m \in M, then m*m^{-1} = 1_M, but since m \in N as well, m*m^{-1} = 1_N. Is that legitimate reasoning? I'm not confident with my answer for some reason.

    (2) Let m,n \in R for a commutative ring R.

    Suppose that mn is a zero divisor in R. Prove that either m or n is a zero divisor.

    So, Well, first what I did was use the defn of zero divisor (z.d.)
    If mn \neq 0 is a z.d. then \exists x \in R such that x \neq 0 and mn(x) = 0

    Then, just by associativity, m(nx) = 0 so, well, can we assume that nx \neq 0 and m \neq 0? Cause we'd be done - m is a z.d. ! By commutativity, just switch m and n and then we're done.

    Any help appreciated! Thanks!!
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  2. #2
    Senior Member roninpro's Avatar
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    Your proof for (1) looks okay. You just need to emphasize that inverses are unique.

    You have a good idea for (2). To clean it up, you should note that neither m nor n should be zero, or else mn is zero, which is not what we want. Then, you are correct in taking nonzero x so that mnx=0 and m(nx)=0. So if nx\ne 0, then you are done, since m will satisfy the zero divisor definition. But what happens if nx=0?
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  3. #3
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    I'd just like to point out that (1) is actually true whenever the rings are integral domains.

    In the ring M, of course 1_M\cdot 1_M=1_M. Also, since 1_M\in N and 1_N is the multiplicative identity of N, we get 1_M\cdot 1_N=1_M.

    Put these equalities together to get 1_M\cdot 1_M=1_M\cdot 1_N. Then, since we are in an integral domain, we can "cancel" the 1_M from both sides to conclude 1_M=1_N.
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